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Algorithms, Distributed System, Machine Learning

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"How many inverted pairs" - that usually ends up with MergeSort solution (of course there are other solutions out there)

def mergeSort(arr):
    if len(arr) == 1:
        return 0, arr
    
    mid = len(arr) // 2
    cnt1, arr1 = mergeSort(arr[:mid])
    cnt2, arr2 = mergeSort(arr[mid:])
    
    #
    ret = []    
    cnt = 0
    
    i = 0
    j = 0
    inx = 0
    while i < len(arr1) and j < len(arr2):
        if arr1[i] <= arr2[j]:
            ret.append(arr1[i])
            i += 1
        elif arr2[j] < arr1[i]:
            ret.append(arr2[j])
            cnt += len(arr1) - i        # arr2 -> arr1
            j += 1        
    while i < len(arr1):
        ret.append(arr1[i])
        i += 1
    while j < len(arr2):
        ret.append(arr2[j])        
        j += 1
    
    return cnt1 + cnt2 + cnt, ret

T = int(input())
for _ in range(0, T):
    n = int(input())
    arr = [int(i) for i in input().strip().split()]

    cnt, _ = mergeSort(arr)
    print (cnt)
posted on 2015-03-18 03:42  Tonix  阅读(328)  评论(0编辑  收藏  举报